1. Find $f(x)$ using the functional equation

Given $f(x)+2f\left(\frac{1}{x}\right)=x^2+5, \qquad x>0$

Replace $x$ by $\frac{1}{x}$: $f\left(\frac{1}{x}\right)+2f(x)=\frac{1}{x^2}+5$

Now let $a=f(x), \qquad b=f\left(\frac{1}{x}\right)$ Then we have the system: $a+2b=x^2+5 \quad ...(1)$ $2a+b=\frac{1}{x^2}+5 \quad ...(2)$

Solve for $a$.

From $(1)$: $a=x^2+5-2b$ Substitute into $(2)$: $2(x^2+5-2b)+b=\frac{1}{x^2}+5$ $2x^2+10-4b+b=\frac{1}{x^2}+5$ $2x^2+10-3b=\frac{1}{x^2}+5$ $3b=2x^2+5-\frac{1}{x^2}$ $b=\frac{2x^2+5-\frac{1}{x^2}}{3}$

Then $a=x^2+5-2b$ $f(x)=x^2+5-\frac{2}{3}\left(2x^2+5-\frac{1}{x^2}\right)$ $f(x)=\frac{3x^2+15-4x^2-10+\frac{2}{x^2}}{3}$ $f(x)=\frac{5-x^2+\frac{2}{x^2}}{3}$

So, $f(x)=\frac{5-x^2+2/x^2}{3}$

2. Compute $\alpha=\int_1^2 f(x)\,dx$

$\alpha=\int_1^2 \frac{5-x^2+2/x^2}{3}\,dx$ $\alpha=\frac{1}{3}\int_1^2 \left(5-x^2+\frac{2}{x^2}\right)dx$

Integrate termwise: $\int 5\,dx=5x, \qquad \int x^2 dx=\frac{x^3}{3}, \qquad \int \frac{2}{x^2}dx=2\int x^{-2}dx=-\frac{2}{x}$

Hence $\alpha=\frac{1}{3}\left[5x-\frac{x^3}{3}-\frac{2}{x}\right]_1^2$

At $x=2$: $10-\frac{8}{3}-1=9-\frac{8}{3}=\frac{19}{3}$

At $x=1$: $5-\frac{1}{3}-2=3-\frac{1}{3}=\frac{8}{3}$

Thus $\alpha=\frac{1}{3}\left(\frac{19}{3}-\frac{8}{3}\right)=\frac{1}{3}\cdot \frac{11}{3}=\frac{11}{9}$

3. Find $g(x)$ using the second functional equation

Given $2g(x)-3g\left(\frac{1}{2}\right)=x, \qquad x>0$

Let $c=g\left(\frac{1}{2}\right)$ Then $2g(x)-3c=x$ $g(x)=\frac{x+3c}{2}$

Now put $x=\frac{1}{2}$: $g\left(\frac{1}{2}\right)=\frac{\frac{1}{2}+3c}{2}$ Since $g\left(\frac{1}{2}\right)=c$, $c=\frac{\frac{1}{2}+3c}{2}$ $2c=\frac{1}{2}+3c$ $-c=\frac{1}{2}$ $c=-\frac{1}{2}$

Therefore $g(x)=\frac{x+3\left(-\frac{1}{2}\right)}{2}=\frac{x-\frac{3}{2}}{2}=\frac{x}{2}-\frac{3}{4}$

4. Compute $\beta=\int_1^2 g(x)\,dx$

$\beta=\int_1^2 \left(\frac{x}{2}-\frac{3}{4}\right)dx$ $\beta=\left[\frac{x^2}{4}-\frac{3x}{4}\right]_1^2$

At $x=2$: $\frac{4}{4}-\frac{6}{4}=1-\frac{3}{2}=-\frac{1}{2}$

At $x=1$: $\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}$

So, $\beta=-\frac{1}{2}-\left(-\frac{1}{2}\right)=0$

5. Compute $9\alpha+\beta$

$9\alpha+\beta=9\left(\frac{11}{9}\right)+0=11$

So the correct option is: $\boxed{\text{D: }11}$

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, the answer agrees with the stored correct answer.